The application of Euler function
Original question:
As a sports Commissioner, C June is responsible for the training of the honor guard. Guard of Honor is composed of students of N * N of the square, in order to ensure that the procession uniform, c June will follow the guard of Honor's left rear, according to the number of students in the line of sight to determine whether the team is neat (such as). Now, C-June wants you to tell him the number of students you can see when the team is neat.
1<=n<=40000
First, the graph along the lower left corner to the upper right corner of the diagonal fold, the two sides are the same, so only one side of the *2+1 can be asked (because there is a diagonal, so + 1)
How do you ask for each side?
You can find that all the points you see meet gcd (x, y) ==1
And then you're asking for 1<=x<=n-1 's Phi.
Why is that?
First y must be smaller than x, or run to the other side (on the face of the folding ↑), and then gcd (x, y) also ==1, is smaller than x and x coprime number of numbers, is also the definition of Euler functions, so directly to Phi and can
It is important to note that the coordinates of the lower left corner should be set to 0 (which is also the reason for the 1<=x<=n-1 Phi and ↑), as this will ensure that X-y is the horizontal distance from the lower-left corner
Code:
1#include <iostream>2#include <cstdio>3#include <algorithm>4#include <cstring>5#include <cmath>6 using namespacestd;7 intN;8 intkang[41000],phi[41000],zhi[410000],ztop=0;9 voidShai () {Tenmemset (Kang,0,sizeof(Kang)); Onephi[1]=1; A for(intI=2; i<=n;i++){ - if(!kang[i]) {phi[i]=i-1; zhi[++ztop]=i;} - for(intj=1; J<=ztop && i*zhi[j]<=n;j++){ thekang[i*zhi[j]]=true; - if(i%zhi[j]==0) {Phi[i*zhi[j]]=phi[i]*zhi[j]; Break;} -phi[i*zhi[j]]=phi[i]*Phi[zhi[j]]; - } + } - } + intMain () {//freopen ("ddd.in", "R", stdin); ACin>>N; at Shai (); - Long LongBowl=0; - for(intI=1; i<n;i++) bowl+=Phi[i]; -cout<<bowl*2+1<<Endl; - return 0; -}
View Code
"SDOI2008" "P1377" honor Guard